Hahn decomposition of measure I'm reading a proof of Hahn decomposition from Bogachev's Measure Theory. I rewrite the proof to enforce my understanding. I'm pleased to receive your suggestions.

Let $X$ be a topological space and $\mathcal B(X)$ its Borel $\sigma$-algebra. Let $\mu$ be a finite signed Borel measure on $X$. A set $B \in \mathcal B(X)$ is called negative (resp. positive) if $\mu(C\cap B) \le 0$ (resp. $\mu(C\cap B) \ge 0$) for all $C \in \mathcal B(X)$.

Theorem: There are $N, P\in \mathcal B(X)$ such that $N \cap P =\emptyset, N \cup P = X$ and that $N,P$ are negative and positive respectively.

I post the proof separately as below answer. This allows me to subsequently remove this question from unanswered list.
 A: Let $\mathcal N$ be the collection of all negative sets. Notice that $\mathcal N$ is closed under countable union and countable intersection. Let $\alpha := \inf \{\mu(B) \mid B \in \mathcal N\}$. Let $(N_n)$ be a sequence in $\mathcal N$ such that $\mu(N_n) \searrow \alpha$. Let $N := \bigcup_n N_n$. Clearly, $N$ is negative and thus $N \in \mathcal N$. We have $\alpha \le \mu(N) \le \mu(N_n)$, so $\mu(N) = \alpha$.
Let $P := X \setminus N$. We will prove that $P$ is a positive. Assume the contrary that $P$ is not positive. Then $\exists B_0 \in \mathcal B(X)$ s.t. $B_0 \subset P$ and $\mu (B_0)<0$. Then $B_0$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0$ and $\mu (B)>0$.

*

*We pick the smallest $k_1 \in \mathbb N$ such that $\exists B_1 \in \mathcal B(X)$ s.t. $B_1 \subset B_0$ and $\mu (B_1) \ge 1/k_1$. Then $\mu(B_0\setminus B_1)<0$. Then $B_0\setminus B_1$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0 \setminus B_1$ and $\mu (B)>0$.


*We pick the smallest $k_2 \in \mathbb N$ such that $\exists B_2 \in \mathcal B(X)$ s.t. $B_2 \subset B_0 \setminus B_1$ and $\mu (B_2) \ge 1/k_2$. Then $\mu(B_0\setminus B_1\setminus B_2)<0$. Then $B_0\setminus B_1 \setminus B_2$ is not negative, so $\exists B \in \mathcal B(X)$ s.t. $B \subset B_0 \setminus B_1 \setminus B_2$ and $\mu (B)>0$.
Inductively, $\exists B_{n+1} \in \mathcal B(X)$ s.t. $B_{n+1} \subset B_0 \setminus B_1 \setminus \cdots \setminus B_n$ and $\mu(B_{n+1})>1/k_{n+1}$. Clearly, $k_n \to \infty$. If not, $\mu(\bigcup_{n=1}^\infty B_n) = \infty$.
Let $C := \bigcup_{n \ge 1} B_n \subset B_0$. Then $\mu(C) >0$. We claim that $B_0 \setminus C$ is negative. If not, $\exists D \in \mathcal B(X)$ s.t. $D \subset B_0 \setminus C$ and $\mu (D)>0$. Take $M \in \mathbb N$ such that $\mu(D) > 1/M$. There is $N \in \mathbb N$ such that
$$
\frac{1}{k_N-1} \le  \frac{1}{k_{N}} + \frac{1}{M} \le \mu(B_N \cup D).
$$
This contradicts the minimality of $k_N$, so $B_0\setminus C$ is negative. This contradicts the minimality of $\alpha$. Hence $P$ is positive.

Let $(N', P')$ be another Hahn decomposition of $\mu$. Let $B \in \mathcal B(X)$. Then
$$
\mu(B\cap N) = \mu(B\cap N\cap N') + \mu(B\cap N \cap P') =   \mu(B\cap N\cap N').
$$
By symmetry, $\mu(B\cap N') = \mu(B\cap N\cap N')$. So $\mu(B\cap N) = \mu(B\cap N')$. Similarly, $\mu(B\cap P) = \mu(B\cap P')$.
A: It may help some readers of the first answer to add some more details.
In the paragraph of that answer beginning with Let P:=X∖N, the conclusion that $B_0$ is not negative may not be obvious. It is obtained like this: if this set were negative, then $\mu(B_0 \cup N) = \mu(B_0) + \mu(N) < \alpha$ because $B_0$ and $N$ are disjoint and $\mu(B_0) < 0$, contradicting the minimality of $\alpha$.
In the next two paragraphs, there are similar conclusions about some set not being negative; these are obtained in the same way.
Then in the paragraph beginning with the definition of $C$, it is true that $\mu(C)>0$ but what really matters here is that $\mu(B_0 \setminus C)<0$. When after that it is shown that $B_0 \setminus C$ is indeed negative, a similar argument shows again a contradiction with the minimality of $\alpha$ (as stated explicitely this time in the 1st answer), this being the last contradiction leading to the conclusion that $P$ is positive (as stated there).
And then there is something else that might confuse a reader of the 1st answer, but which a smart one will be able to correct himself. It is the fact that the letter $N$ is used in two different contexts with different roles. First it is a set as constructed in the first paragraph of the proof. Towards the end, it is a natural number - which by the way must be non-zero (no $k_0$ has been defined).
Hope that helps.
